Answer:
The marble was moving in a projectile and the speed of the engine was 2.716 m/s
Explanation:
The vertical component of the marble's flight path relative to the train
is given by the equation y(t) = v*t - (4.9)*t^2,
where, v is the initial upward velocity of the marble relative to the train.
So with y(1) = v - 4.9 = 0 we have
v = 4.9 m/s.
The marble will reach maximum height after 0.5 seconds, at which the
height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.
Now, the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component
of V m/s such that tan(61) = 4.9 / V
V = 4.9 / tan(61) = 2.716 m/s
This horizontal velocity component of the marble is the same as the
speed of the train i.e. 2.716 m/s.
<span>A = area of styrofoam
M = mass of stryofoam = A*h*rho_s
m = mass of swimmer
Total mass = m + M = m + A*h*rho_s
Downward force = g*(total mass) = g*[m + A*h*rho_s]
The slab is completely submerged.
Buoyant force = g*(mass of water displaced) = g*[A*h*rho_w]
Equate these
g*[m + A*h*rho_s] = g*[A*h*rho_w]
m + A*h*rho_s = A*h*rho_w
A*h*[rho_w - rho_s] = m
A = m/[h*(rho_w - rho_s)]</span>
B.
The child is too old to be gaining something from the screen time.
Answer:
The distribution is as depicted in the attached figure.
Explanation:
From the given data
- The plane wall is initially with constant properties is initially at a uniform temperature, To.
- Suddenly the surface x=L is exposed to convection process such that T∞>To.
- The other surface x=0 is maintained at To
- Uniform volumetric heating q' such that the steady state temperature exceeds T∞.
Assumptions which are valid are
- There is only conduction in 1-D.
- The system bears constant properties.
- The volumetric heat generation is uniform
From the given data, the condition are as follows
<u>Initial Condition</u>
At t≤0
This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.
<u>Boundary Conditions</u>
<u>At x=0</u>
<u />
<u />
This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.
<u>At x=L</u>
<u />![-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]](https://tex.z-dn.net/?f=-k%5Cfrac%7B%5Cpartial%20T%7D%7B%5Cpartial%20x%7D%5D_%7Bx%3DL%7D%3Dh%5BT%28L%2Ct%29-T_%7B%5Cinfty%7D%5D)
<u />
This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.
The temperature distribution along with the schematics are given in the attached figure.
Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.
It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.
:<span> </span><span>30.50 km/h = 30.50^3 m / 3600s = 8.47 m/s
At the top of the circle the centripetal force (mv²/R) comes from the car's weight (mg)
So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road ..
Fn = mg - mv²/R
Fn = m(g - v²/R) .. .. 1800kg (9.80 - 8.47²/20.20) .. .. .. ►Fn = 11 247 N (upwards)
(b)
When the car's speed is such that all the weight is needed for the centripetal force .. then the net downward force (Fn), and the reaction from the road, becomes zero.
ie .. mg = mv²/R .. .. v² = Rg .. .. 20.20m x 9.80 = 198.0(m/s)²
►v = √198 = 14.0 m/s</span>
